How can I show a function that converts a tuple to a non-tuple is bijective?

Question

I have the below function I came up with and would like to show that it is injective and surjective (bijective). How can I do this?

$$f\bigg(\bigg(\frac{a}{b}, \frac{c}{d}\bigg)\bigg) = (a+c)+\sqrt7(b+d)$$

To prove the function is injective, I tried supposing $\ f\bigg(\bigg(\frac{a_1}{b_1}, \frac{c_1}{d_1}\bigg)\bigg) = \ f\bigg(\bigg(\frac{a_2}{b_2}, \frac{c_2}{d_2}\bigg)\bigg)$ and then showing that $(a_1+c_1)+\sqrt7(b_1+d_1) = (a_2+c_2)+\sqrt7(b_2+d_2)$.

I am not sure if this is the correct way to go about this.

EDIT: Domain is $\mathbb{Q} \times \mathbb{Q}$, range is \mathbb{R}$.

Answer 1

To show Injectivity it suffices to show that $\forall x,y \in Dom(f)$ if $f(x)=f(y) \implies x=y$. However, keep in mind that the choice of $x$ and $y$ must be arbitrary. In your solution, it does seem to be injective.

In order to show surjectivity you have to show $\forall y\in Ran(f), \exists x\in Dom(b)$. However, you must specify $f:A \to B$. That is, what kind of map is $f$, from what set to what set. Is it from $\mathbb R \to \mathbb R, \mathbb R \to \mathbb Z$, etc.

Answer 2

There is a problem with f. Let r = $\sqrt7$.
2 + 2r = f(1/1,1/1) = f(2/2,1/1) = 3 + 3r.

It is not surjective because the range of f is a subset of the algebraic numbers and $\pi$ is not algebraic. Thus the codomain of f cannot be R.

To fix the problem, require a,b,c,d to be integers with positive b,d and a/b, c/d to be in lowest terms. Now f is well defined.

Its range is { n + mr : n integer, m positive integer }.
It is not injective because f(a/b,c/d) = f(c/d,a/b).